## Math 9 Review chapter 4: Quadratic equation one unknown

## 1. Summary of theory

### 1.1. Graph of the function \(y=ax^2 (a\neq 0)\)

The graph of the function \(y=ax^2 (a\neq 0)\) is the set of all points \(M(x_{M}; ax_{M}^{2})\). To determine a point on the graph, we take a value of x as the coordinate and substitute the equation \(y=ax^2\) to find the coordinate value.

### 1.2. Quadratic

An unknown quadratic equation (referred to as a quadratic equation) is an equation of the form \(ax^2+bx+c=0\)

Where, x is hidden; the coefficients a, b, c are given numbers and \(a\neq 0\)

### 1.3. Solution formula of quadratic equation

With the equation \(ax^2+bx+c=0 (a\neq 0)\) and the differential \(\Delta =b^2-4ac\):

- \(\Delta>0\) then the equation has 2 distinct solutions:
- \(x_{1}=\frac{-b+\sqrt{\Delta }}{2a}\); \(x_{2}=\frac{-b-\sqrt{\Delta }}{2a}\)
- \(\Delta=0\) then the equation has a double solution \(x=x_{1}=x_{2}=-\frac{b}{2a}\)
- \(\Delta<0\) the equation has no solution.

### 1.4. Compact test formula

For the quadratic equations \(ax^2+bx+c=0(a\neq 0)\) and \(b=2b’\), \(\Delta ‘=b’^2-ac\) then :

- If \(\Delta ‘>0\) then the equation has two distinct solutions: \(x_{1}=\frac{-b’+\sqrt{\Delta ‘}}{a}; x_{2}= \frac{-b’-\sqrt{\Delta ‘}}{a}\)
- If \(\Delta ‘=0\) then the equation has a double solution \(x=\frac{-b’}{a}\)
- If \(\Delta ‘<0\) then the equation has no solution.

### 1.5. Viet’s theorem and its applications

The quadratic equation \(ax^2+bx+c=0(a\neq 0)\) has 2 distinct solutions

\(x_1=\frac{-b+\sqrt{\Delta }}{2a}; x_2=\frac{-b-\sqrt{\Delta }}{2a}\)

We have: \(x_1+x_2=\frac{-2b+\sqrt{\Delta }-\sqrt{\Delta }}{2a}=-\frac{b}{a}\)

\(x_1.x_2=\frac{b^2-\Delta }{4a^2}=\frac{4ac}{4a^2}=\frac{c}{a}\)

**Viet’s theorem**

If \(x_1;x_2\) are two solutions of the equation \(ax^2+bx+c=0 (a\neq 0)\) then:

\(x_1+x_2=-\frac{b}{a}\) and \(x_1.x_2=\frac{c}{a}\)

**generality**

- If the equation \(ax^2+bx+c=0 (a\neq 0)\) has \(a+b+c=0\) then the equation has a solution of \(x_1=1\) and a solution. the other is \(x_2=\frac{c}{a}\).
- If the equation \(ax^2+bx+c=0 (a\neq 0)\) has \(a-b+c=0\) then the equation has a solution of \(x_1=-1\) and the other solution is \(x_2=-\frac{c}{a}\).

### 1.6. Equations that are reduced to quadratic equations (quadratic equations, equations with hidden samples, product equations…)

**a. The quadratic equation**

**Define**

A quadratic equation is an equation of the form: \(ax^4+bx^2+c=0 (a\neq 0)\)

**b. Equation hidden in the sample**

Steps to solve the equation containing the implicit in the sample learned in grade 8

- Step 1: Find the definite condition of the equation
- Step 2: Convert both sides and then de-sample
- Step 3: Solve the equation you just got
- Step 4: Compare the initial conditions and then conclude the solution

**c. Productivity Equation**

Repeat what you learned in class below:

Convert the equation to the form \(ABC….=0\) and then deduce either \(A=0\) or \(B=0\) or…..

### 1.7. Solve math problems using the equation method

**Solution method**

To solve a problem by making an equation, we follow these steps:

__ Step 1:__ Equation

- Select hide and set conditions for hiding
- Express different quantities in implicit terms
- Based on the problem, make an equation in the form you have learned

__ Step 2:__ Solve the equation

__ Step 3:__ Compare the results and choose the appropriate solution

## 2. Illustrated exercise

### 2.1. Exercise 1

Given the function \(y=-x^2\) and the line \(y=-4x+4\). Find the intersection of those two graphs by drawing and graph

**Solution guide**

Draw a self-drawn picture.

Find intersection: Equation of coordinates of intersection: \(-x^2=-4x+4\Leftrightarrow x^2-4x+4=0\)

The discriminant \(\Delta=0\) derives the equation with a double solution \(x=2\)

So when we draw the figure, we only get one intersection. Later on, they will know that the above line is a tangent to the function.

### 2.2. Exercise 2

Solve the equation by factoring the polynomial: \(x^2-11x-12=0\)

**Solution guide**

\(x^2-11x-12=0\)

\(\Leftrightarrow x^2-12x+x-12=0\)

\(\Leftrightarrow x(x-12)+x-12=0\)

\(\Leftrightarrow (x+1)(x-12)=0\)

So the above equation has two distinct solutions, \(x=-1;x=12\)

### 2.3. Exercise 3

Solve the equation: \(x^2+10x+25=0\); \(x^2-4x-9=0\)

**Solution guide**

+ \(x^2+10x+25=0\)

\(\Delta =10^2-4.1.25=0\) \(\Rightarrow x=\frac{-0}{2}=-5\)

+ \(x^2-4x-9=0\)

\(\Delta =(-4)^2-4.1.(-9)=52\Rightarrow \sqrt{\Delta }=2\sqrt{13}>0\)

\(\Rightarrow x_{1}=\frac{-(-4)+2\sqrt{13}}{2}=2+\sqrt{13};x_{2}=\frac{-(-4) -2\sqrt{13}}{2}=2-\sqrt{13}\)

### 2.4. Exercise 4

Find two numbers whose difference is 5 and their product is 150

**Solution guide**

Let’s call the two numbers that need to be found a, b

We have \(\left\{\begin{matrix} ab=5\\ ab=150 \end{matrix}\right.\)

Putting \(a=5+b\) into the product equation, we get \(b(b+5)=150\Leftrightarrow b^2+5b-150=0\)

\(\Rightarrow b=-15\) or \(b=10\)

\(b=-15\Rightarrow a=-10\)

\(b=10\Rightarrow a=15\)

### 2.5. Exercise 5

Solve the following quadratic equation: \(x^4-4x^2-5=0\)

__Solution guide__

Set \(t=x^2 (t\geq 0)\)

Then the equation becomes: \(t^2-4t-5=0\)

Solving the above quadratic equation, we get:

\(t=-1\) (type)

\(t=5\) (get)\(\Rightarrow x=\pm \sqrt{5}\)

## 3. Practice

### 3.1. Essay exercises

**Question 1:** Given two functions: \(y = 2x – 3\) and \(y = – {x^2}\)

a) Plot the graphs of these two functions in the same coordinate plane.

b) Find the coordinates of the intersection points of the two graphs.

Test that the coordinates of each intersection are common solutions of the two equations \(y = 2x – 3 \) and \(y = – {x^2}\)

**Verse 2: **Solve the equations:

a) \(3{x^2} + 4\left( {x – 1} \right) = {\left( {x – 1} \right)^2} + 3\)

b) \({x^2} + x + \sqrt 3 = \sqrt 3 x + 6\)

c) \(\displaystyle{{x + 2} \over {1 – x}} = {{4{x^2} – 11x – 2} \over {\left( {x + 2} \right)\left ( {x – 1} \right)}}\)

d) \(\displaystyle{{{x^2} + 14x} \over {{x^3} + 8}} = {x \over {x + 2}}\)

**Question 3: **Solving quadratic equations

a) \({x^4} + 2{x^2} – x + 1 = 15{x^2} – x – 35\)

b) \(2{x^4} + {x^2} – 3 = {x^4} + 6{x^2} + 3\)

c) \(3{x^4} – 6{x^2} = 0\)

d) \(5{x^4} – 7{x^2} – 2 = 3{x^4} – 10{x^2} – 3\)

**Question 4: **Solve the following equations using the sub-hidden method:

a) \({\left( {{x^2} – 2x} \right)^2} – 2{x^2} + 4x – 3 = 0\)

b) \(3\sqrt {{x^2} + x + 1} – x = {x^2} + 3\)

### 3.2. Multiple choice exercises

**Question 1:** The solution set of the equation \(x^2 < 100\) is:

A. \(x>-10\)

B. \(x>10\)

C. \(x<10\)

D. \(-10

**Verse 2: **The sum and product of 2 solutions of the equation \(x^2+2016x-2017=0\) are respectively:

A. \(S=-2016;P=-2017\)

B. \(S=2016;P=2017\)

C. \(S=-2016;P=2017\)

D. \(S=2016;P=-2017\)

**Question 3:** The coordinates of the intersection of the functional equation \(y=x^2\) and the line \(y=8\) are:

A. \((0,0)\)

B. \((2\sqrt{2};0)\)

C. \((-2\sqrt{2};0)\)

D. \((-2\sqrt{2};0)(2\sqrt{2};0)\)

**Question 4:** For what value of m does the quadratic equation \(x^2+6x-m=0\) have no solution?

A. \(m<-9\)

B. \(m>-9\)

C. \(m>9\)

D. \(m<9\)

**Question 5:** Find two numbers whose sum is 30 and sum of squares is 468.

A. \(14;16\)

B. \(17;13\)

C. \(18;12\)

D. \(19;11\)

## 4. Conclusion

Through this lesson, students need to:

- Have skills in making analytical tables, representing quantities in the problem according to the given unknowns and known quantities.
- Apply enough steps to complete the problem by making equations
- Apply to solve some basic maths, not too complicated.
- Do real problems to see clearly that mathematics originates from real life and comes back to serve reality.

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